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How to prove this inequation?

$\displaystyle\left(\sum_{j=1}^{N}a_j\right)^{\theta}\leq N^{\theta-1}\displaystyle\sum_{j=1}^{N}{a_j}^{\theta}$,

where $a_j$ be a sequence of positive reals and $1 ≤ θ < ∞$.

I try to rewrite this inequation into $1≤N^{θ-1}\left[\left(\frac{a_1}{\sum_{j=1}^{N}{a_j}}\right)^θ+\left(\frac{a_2}{\sum_{j=1}^{N}{a_j}}\right)^θ+...+\left(\frac{a_n}{\sum_{j=1}^{N}{a_j}}\right)^θ\right]$. But what should I do next? For $N=2$, I can prove it.

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    $\begingroup$ Another way would be to use the convexity of $t\mapsto t^\theta$ for $\theta\geq1$. $\endgroup$ – Jose27 Jun 6 '17 at 5:57
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This follows from the generalisation of Titu's Lemma (which is a consequence of Cauchy-Schwarz and Holder inequalities), where if $a_i,b_i>0$, and $\theta\geq1$ we have $$\sum_{i = 1}^N \frac{a_i^\theta}{b_i^{\theta-1}}\geq\frac{\left(\sum_{i=1}^Na_i\right)^\theta}{\left(\sum_{i=1}^Nb_i\right)^{\theta-1}}$$ In your case, set $b_i=1$ for all $i$ and you get your desired result: $$\sum_{i = 1}^N a_i^\theta\geq\frac{\left(\sum_{i=1}^Na_i\right)^\theta}{N^{\theta-1}} \implies \left(\sum_{i=1}^Na_i\right)^\theta\leq N^{\theta-1}\cdot\sum_{i = 1}^N a_i^\theta$$

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Hint:

$$ \begin{align} \left(\sum_{j=1}^{N}a_j\right)^{\theta}\leq N^{θ-1}\displaystyle\sum_{j=1}^{N}{a_j}^{θ} \quad &\iff\quad \left(\frac{\sum_{j=1}^{N}a_j}{N}\right)^{\theta} \leq \frac{\sum_{j=1}^{N}{a_j}^{θ}}{N} \\ &\iff\quad \frac{\sum_{j=1}^{N}a_j}{N} \leq \left(\frac{\sum_{j=1}^{N}{a_j}^{θ}}{N}\right)^{\frac{1}{\theta}} \end{align} $$

The latter is just the generalized mean inequality for $\theta \ge 1\,$, discussed for example here.

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It's just Power Mean inequality. $$\left(\frac{\sum\limits_{j=1}^{N}a_j}{N}\right)^{\theta}\leq \frac{\sum\limits_{j=1}^{N}a_j^{\theta}}{N}$$

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  • $\begingroup$ Power Mean inequality A.k.a. the generalized mean inequality, as I quoted it. $\endgroup$ – dxiv Jun 6 '17 at 6:41
  • $\begingroup$ @dxiv It's so obvious, that I don't know what to say. It's obvious also by Jensen and more... $\endgroup$ – Michael Rozenberg Jun 6 '17 at 6:44
  • $\begingroup$ I initially had just a comment to that effect, but then posted it as an answer because of a mixup with OP's question (see the edit history of both the posted question and my answer if curious). $\endgroup$ – dxiv Jun 6 '17 at 6:49
  • $\begingroup$ @dxiv Indeed, our answers are very similar. What do you want that I'll do? $\endgroup$ – Michael Rozenberg Jun 6 '17 at 8:24
  • $\begingroup$ Just leave it as-is. Overlaps like these will happen, especially in cases of obvious like you said. $\endgroup$ – dxiv Jun 6 '17 at 17:09

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